Interferometer to the Horizontal Pendulum. 507 



distance it would require two equal masses m (since 7 == 6 - 7 X 

 10~ 8 roughly) of 180 grams, per vanishing interference ring, 

 at a distance of 1 centimeter. These conditions could be sup- 

 plied by using parallel plates. On the other hand, the frame- 

 work of the above pendulum is unnecessarily heavy, and was 

 constructed out of the material at hand. It could easily be 

 reduced in weight much below the above datum, or the 

 greater part supported on a float, so that the case is to be 

 stated many times more favorably. 



Resuming equation (3), if K is the moment of inertia, i the 

 radius of gyration, and T the period and / the length of the 

 horizontal pendulum, 



/" 



T =^V^ (•) 



an equation from which <£ may be found in terms of T, i, and 

 h, which must be measured. 



Again the indicated length H of the pendulum (distance 

 from the center of gravity to the point of intersection of the 

 axis and the plumbline through the center of gravity) is 



H= A/sin </> = h/cji, nearly. (7) 



The change of vertical inclination a of the axis of the pen- 

 dulum corresponding to the horizontal deviation is then, 

 nearly, 



a = yr = 005 nearly; (8) 



or if the period T be introduced from (6) and 6 from (5) 



_ 4ttV MV 



It is in equation (8) that the condition of remarkable sensitive- 

 ness resides. Thus, if the interferometer is used, 



and, if AiV = 30 X 10" 6 and <j> = 10" 2 (somewhat less than 

 1° of arc), R = 111 cm., as above, 



a = 13 X 10" 1 ° radians = -00028" 



per vanishing interference ring. 



If F, as before, is the force at the center of gravity, the 

 corresponding force at the grating, a distance R from the 

 center, is 



F R =Mg$d^=M^6 (10) 



since (f> is given by equation (6). 



